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Re:拓扑排序后 dp[a][a]=1 dp[a][b]=dp[b][a]=sum(dp[a][parent[b]])/2 when order(a)>order(b)

Posted by lddlinan at 2020-06-30 19:43:42 on Problem 1202
In Reply To:拓扑排序后 dp[a][a]=1 dp[a][b]=dp[b][a]=sum(dp[a][parent[b]]) when order(a)>order(b) Posted by:lddlinan at 2020-06-30 19:23:48


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